Help needed

general
1

I am a beginner level coder. I’m stuck at this problem Courses - Codeforces
and here is my code
my_code - Pastebin.com
Please let me know where I am doing wrong

2

The upper bound for the answer can be as large as 2 \times10^9. Because

N\le 2\times10^8 \ \text{and }\ x \le10 \implies N\times x \le 2 \times 10^9
This should pass ig haven't tested though 
#include <bits/stdc++.h>
using namespace std;
 
typedef vector<int> vi;
typedef pair<int, int> pii;
#define endl "\n"
#define sd(val) scanf("%d", &val)
#define ss(val) scanf("%s", &val)
#define sl(val) scanf("%lld", &val)
#define debug(val) printf("check%d\n", val)
#define all(v) v.begin(), v.end()
#define pb push_back
#define mp make_pair
#define ff first
#define Ss second
#define ll long long
#define ull unsigned long long
#define MOD 1000000007
#define clr(val) memset(val, 0, sizeof(val))
#define what_is(x) cerr << #x << " is " << x << endl;
#define OJ                            \
    freopen("input.txt", "r", stdin); \
    freopen("output.txt", "w", stdout);
#define FIO                           \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
 
ll n;
ll x, y;
bool good(ll p)
{
    ll m = ((p - min(x, y)) / max(x, y)) + (p / min(x, y));
 
    return m >= n;
}
 
int main()
{
 
    cin >> n >> x >> y;
 
    ll l = 0;
    ll r = 2e9 ;
 
    while (l + 1 < r)
    {
        ll mid = (l + r) / 2;
        if (good(mid))
        {
            r = mid;
        }
        else
            l = mid;
    }
 
    cout << r << endl;
 
    return 0;
}
3

Here is the code

4

Yeah ! it did solve my problem. Thanks

5

I’ve learnt certain things, thanks